Pedant
07-08-2003, 04:25 AM
(all calculations and figures approximate)
shc air 100 celcius = 1000 (J/(kg * degree K)
1kg air = ~1 metre cubed
one cubic metre per minute = 35 cfm
100 watts for one minute = 6000 Joules of energy to be dissipated.
6 celsius rise in air temperature (for 1 metre cubed with 6000 Joules of heat added to it)
Therefore, with a CPU releasing 100 watts of heat, a heatsink with perfect transfer of heat between it and the air (i.e. a temperature gradient of 0 between it and the air, as well as across it), and a fan pushing 35cfm, the heatsink would be at six degrees above ambient.
shc air 100 celcius = 1000 (J/(kg * degree K)
1kg air = ~1 metre cubed
one cubic metre per minute = 35 cfm
100 watts for one minute = 6000 Joules of energy to be dissipated.
6 celsius rise in air temperature (for 1 metre cubed with 6000 Joules of heat added to it)
Therefore, with a CPU releasing 100 watts of heat, a heatsink with perfect transfer of heat between it and the air (i.e. a temperature gradient of 0 between it and the air, as well as across it), and a fan pushing 35cfm, the heatsink would be at six degrees above ambient.